We notice that the denominator of each ratio is the product of 2 consecutive numbers.

We can write this kind of ratio 1/n(n+1) as the result of addition or subtraction of 2 elementary fractions:

We'll write the ratio:

1/n(n+1) = A/n + B/(n+1) (1)

We'll multiply the ratio A/n by (n+1) and we'll multiply the ratio B/(n+1) by n.

1/n(n+1) = [A(n+1) + Bn]/n(n+1)

Since the denominators of both sides are matching, we'll write the numerators, only.

1 = A(n+1) + Bn

We'll remove the brackets:

1 = An + A + Bn

We'll factorize by n to the right side:

1 = n(A+B) + A

If the expressions from both sides are equivalent, the correspondent coefficients are equal.

A+B = 0

A = 1

1 + B = 0

B = -1

We'll substitute A and B into the expression (1):

**1/n(n+1) = 1/n - 1/(n+1) (2)**

According to (2), we'll give values to n:

for n = 1 => 1/1*2 = 1/1 - 1/2

for n = 2 => 1/2*3 = 1/2 - 1/3

.....................................................

for n = n-1 => 1/n(n-1) = 1/(n-1) - 1/n

for n = n => 1/n(n+1) = 1/n - 1/(n+1)

If we'll add the ratios from the left side, we'll get:

1/1*2 + 1/2*3 + ... + 1/n(n-1) + 1/n(n+1) = 1/1 - 1/2 + 1/2 - 1/3 + ..... + 1/(n-1) - 1/n + 1/n - 1/(n+1)

We'll eliminate like terms from the right side:

1/1*2 + 1/2*3 + ... + 1/n(n-1) + 1/n(n+1) = 1 - 1/(n+1)

S = (n + 1 - 1)/(n+1)

We'll eliminate like terms and we'll get the final result of the sum:

**S = n /(n+1)**

To find the sum 1/(1*2) +1/(2*3)+1/(3*4)+....1/n(n+1).

Let Sn = 1/(1*2) +1/(2*3)+1/(3*4)+....1/n(n+1).

Consider 1/n(n+1) = 1/n -1/(n+1)

Threfore we can split each term of the given series as below:

Sn = {(1/1-1/2) + (1/2 -1/3) +(1/3-1/4) + (1/4-1/5) +..... +(1/n-1/(n+1)}

Sn = 1-1/2 +1/2 -1/3 +1/3 -1/4 +1/4 +......-/n+1/n -1/(n+1).

Sn = 1 +0 +0+...+0-1/(n+1)

Therefore Sn = 1-1/(n+1)

Or Sn = (n+1-1)/(n+10.

Sn = n/(n+1).

Therefore the sum to n terms of 1/(1*2) +1/(2*3)+1/(3.4)+....1/n(n+1) is equal to **n/(n+1).**